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Homework Help: Infinite Series help needed

  1. Feb 28, 2010 #1
    http://www4a.wolframalpha.com/Calculate/MSP/MSP167199e5bhg1gg5673i000048ed16i8cbf5iacg?MSPStoreType=image/gif&s=14&w=256&h=40 [Broken]

    for all x in the interval of convergence of the given power series.

    a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www4a.wolframalpha.com/Calculate/MSP/MSP34199e5dhc325d3hd200005ag14201e835f640?MSPStoreType=image/gif&s=52&w=70&h=36 [Broken]

    b) Find the sum of the series as determined in part a.

    I have no idea how to do these two parts. I just don't know where to begin. Any help is appreciated thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 28, 2010 #2

    rock.freak667

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    Your general term is

    [tex]\frac{(n+1)x^n}{3^{n+1}}[/tex]

    so your sum can be represented as

    [tex]f(x) = \sum_{n=0} ^N \frac{(n+1)x^n}{3^{n+1}}[/tex]

    Now when you integrate within the summation sign, you just integrate what is being summed

    e.g.

    [tex] \int \sum_{n=0} ^N f(x) dx = \sum_{n=0} ^N \int f(x) dx[/tex]
     
  4. Feb 28, 2010 #3
    Okay so I can just take the integral of the general term and find the sum of the first 3 nonzero terms?
     
  5. Feb 28, 2010 #4

    rock.freak667

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    Yes, then you can write out the first three terms and easily find a closed form for the sum.
     
  6. Feb 28, 2010 #5
    Ok, I understand that, but what exactly are the limits of integration (0,1) there for?
     
  7. Feb 28, 2010 #6

    rock.freak667

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    Because you will need to integrate between those limits

    [tex]\sum_{n=0} ^N \int_0 ^1 f(x) dx[/tex]
     
  8. Feb 28, 2010 #7
    Agh! Ok I'm confused. If I'm just taking the integral of the general form, do I still need those limits? Or am I doing a Fundamental Theorem thing?
     
  9. Feb 28, 2010 #8

    rock.freak667

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    yes you would still need those limits. It would be equivalent of doing the integral between those limits of every term.


    Thus you just need to compute

    [tex]\int_0 ^1 \frac{(n+1)x^n}{3^{n+1}}dx [/tex]
     
  10. Feb 28, 2010 #9
    Okay so how exactly would I integrate this? Seeing as there are two variables.
     
  11. Feb 28, 2010 #10

    rock.freak667

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    Treat 'n' as a constant.
     
  12. Feb 28, 2010 #11
    Ok so for the integral I got http://www4a.wolframalpha.com/Calculate/MSP/MSP758199e8ha121cf0a4c00004ggc11fi1gf83a0h?MSPStoreType=image/gif&s=29&w=34&h=40 [Broken]

    Is that right?

    and from that is the sum: infinity?
     
    Last edited by a moderator: May 4, 2017
  13. Feb 28, 2010 #12

    rock.freak667

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    substitute the limits of the integral.


    But now your sum will go from n=-1 to ∞
     
    Last edited by a moderator: May 4, 2017
  14. Feb 28, 2010 #13
    Okay with the limits of integration substituted I got http://www2.wolframalpha.com/Calculate/MSP/MSP344199e931ddgafc99d0000507h04h0277efcda?MSPStoreType=image/gif&s=9&w=34&h=37

    and then i find the first 3 nozero terms from that right?
     
    Last edited by a moderator: Apr 24, 2017
  15. Feb 28, 2010 #14
    Now, I can't see the original question (the link to wolframalpha is dead), but this doesn't sound right.

    Can you post the problem again, sobek?
     
  16. Feb 28, 2010 #15
    http://www2.wolframalpha.com/Calculate/MSP/MSP608199e96816fc40d8a000022aig2805fea6cad?MSPStoreType=image/gif&s=10&w=65&h=38

    for all x in the interval of convergence of the given power series.

    a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www2.wolframalpha.com/Calculate/MSP/MSP179199e974h5cca8ib200004hf8af3ab6c4i576?MSPStoreType=image/gif&s=12&w=70&h=36

    b) Find the sum of the series as determined in part a.

    I'm still having trouble on part A
     
    Last edited by a moderator: Apr 24, 2017
  17. Feb 28, 2010 #16
    Okay, I see it now.

    Yes.

    Assuming your original power series started at n=0, start from there.

    Let n=0 in [tex]\frac 1{3^{n+1}}[/tex] to find the first term. Then let n=1 to find your next, etc.
     
    Last edited by a moderator: Apr 24, 2017
  18. Feb 28, 2010 #17
    Okay so my math in getting https://www.physicsforums.com/latex_images/26/2602939-0.png [Broken] was right?

    and because it's a geometric series I can use a/1-r to find the sum right?
     
    Last edited by a moderator: May 4, 2017
  19. Feb 28, 2010 #18
    Yes and yes.
     
  20. Feb 28, 2010 #19
    Thanks to rocknerd667 and mathnerdmo for helping me with this problem
     
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